如何将NSInteger转换为NSString数据类型？

Question

How does one convert NSInteger to the NSString datatype?

I tried the following, where month is an NSInteger :

  NSString *inStr = [NSString stringWithFormat:@"%d", [month intValue]];

Answer 1

NSIntegers不是对象，您将它们转换为long ，以匹配当前的64位体系结构的定义：

NSString *inStr = [NSString stringWithFormat: @"%ld", (long)month];

Answer 2

Obj-C方式=）：

NSString *inStr = [@(month) stringValue];

Answer 3

Modern Objective-C

An NSInteger has the method stringValue that can be used even with a literal

NSString *integerAsString1 = [@12 stringValue];

NSInteger number = 13;
NSString *integerAsString2 = [@(number) stringValue];

Very simple. Isn't it?

Swift

var integerAsString = String(integer)

Answer 4

%zd works for NSIntegers ( %tu for NSUInteger) with no casts and no warnings on both 32-bit and 64-bit architectures. I have no idea why this is not the " recommended way ".

NSString *string = [NSString stringWithFormat:@"%zd", month];

If you're interested in why this works see this question .

Answer 5

Easy way to do:

NSInteger value = x;
NSString *string = [@(value) stringValue];

Here the @(value) converts the given NSInteger to an NSNumber object for which you can call the required function, stringValue .

Answer 6

在编译支持arm64 ，这不会产生警告：

[NSString stringWithFormat:@"%lu", (unsigned long)myNSUInteger];

Answer 7

You can also try:

NSInteger month = 1;
NSString *inStr = [NSString stringWithFormat: @"%ld", month];

Answer 8

The answer is given but think that for some situation this will be also interesting way to get string from NSInteger

NSInteger value = 12;
NSString * string = [NSString stringWithFormat:@"%0.0f", (float)value];

Answer 9

NSNumber may be good for you in this case.

NSString *inStr = [NSString stringWithFormat:@"%d", 
                    [NSNumber numberWithInteger:[month intValue]]];